# How to solve a differential equation which is not a linear constant coefficient [closed]

How to obtain the general solution of this below Linear differential equation.

\begin{align*} \frac{d^2}{dx^2}\left(\left(1-\frac{a}{b}x\right)^4\frac{d^2Y(x)}{dx^2}\right)-\frac{\omega^2}{c b^2}\left(1-\frac{a}{b}x\right)^2 Y(x)=0 \end{align*}

## closed as off-topic by Kavi Rama Murthy, Arnaud D., Adrian Keister, user10354138, NamasteNov 28 '18 at 15:41

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## 1 Answer

Unless someone knows some very powerful technique, I doubt anyone is going to work this out for you. But I would let $$P(x)=1-\frac{a}{b}x$$ and $$q=\frac{\omega^2} {cb^2}$$ and re-write the equation in the form

$$\left(P^4(x)Y^{\prime\prime}\right)^{\prime\prime}-qP^2(x)Y(x)=0$$

Take the second derivative of the parenthetical term on the left to obtain a fourth order linear equation with polynomial coefficients.

Then you can try solving by various methods such as

1. finding a power series solution on some interval about zero (depending on the roots of the leading polynomial coefficient)
2. finding the Laplace transform of the solution

ADDENDUM: Let $$Y(x)=Y\left(\frac{b}{a}(1-P)\right)=G(P)$$ then I think you will get a Cauchy-Euler equation in $$G$$ and $$P$$.

• Actually, the journal solution is reported in a journal paper, I am just curious how they have obtained it. I tried in Mathematica but could not able to get the answer. I am trying to reproduce the journal paper results. – musimathics Nov 28 '18 at 8:58
• Hint: I believe you can turn this into a Cauchy-Euler equation. – John Wayland Bales Nov 28 '18 at 9:10
• Ok I will try that – musimathics Nov 28 '18 at 9:13
• Let $Y(x)=Y\left(\frac{b}{a}(1-P)\right)=G(P)$ then you get a Cauchy-Euler equation in $G$ and $P$. I will add this to my answer. – John Wayland Bales Nov 28 '18 at 9:17
• Thank you, mean while I will also try. – musimathics Nov 28 '18 at 9:20